`C_6H_6` freezes at `5.5^(@)C`. The temperature at which a solution 10 g of `C_4H_(10)` in 200g of `C_6H_6` freeze is __________`""^(@)C`. (The molal freezing point depression constant of `C_6H_6` is `5.12^(@)C//m`.)

Answer

To solve the problem, we need to determine the freezing point of a solution made by dissolving 10 g of butane (C₄H₁₀) in 200 g of benzene (C₆H₆). We will use the formula for freezing point depression.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of butane (solute), \( m_{solute} = 10 \, g \)
- Mass of benzene (solvent), \( m_{solvent} = 200 \, g = 0.2 \, kg \)
- Freezing point of pure benzene, \( T_f^{\text{benzene}} = 5.5 \, °C \)
- Freezing point depression constant of benzene, \( K_f = 5.12 \, °C/m \)

2. **Calculate the Molar Mass of Butane (C₄H₁₀):**
- Molar mass of C = 12 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of butane = \( (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \, g/mol \)

3. **Calculate the Number of Moles of Butane:**
\[ n_{solute} = \frac{m_{solute}}{M_{solute}} = \frac{10 \, g}{58 \, g/mol} \approx 0.1724 \, mol \]

4. **Calculate the Molality of the Solution:**
\[ m = \frac{n_{solute}}{m_{solvent}} = \frac{0.1724 \, mol}{0.2 \, kg} = 0.862 \, mol/kg \]

5. **Calculate the Freezing Point Depression (\( \Delta T_f \)):**
\[ \Delta T_f = K_f \times m = 5.12 \, °C/m \times 0.862 \, mol/kg \approx 4.41 \, °C \]

6. **Calculate the Freezing Point of the Solution:**
\[ T_f^{\text{solution}} = T_f^{\text{benzene}} - \Delta T_f = 5.5 \, °C - 4.41 \, °C \approx 1.09 \, °C \]

### Final Answer:
The temperature at which the solution freezes is approximately **1.09 °C**.